Petkovšek's algorithm

Petkovšek's algorithm (also Hyper) is a computer algebra algorithm that computes a basis of hypergeometric terms solution of its input linear recurrence equation with polynomial coefficients. Equivalently, it computes a first order right factor of linear difference operators with polynomial coefficients. This algorithm was developed by Marko Petkovšek in his PhD-thesis 1992.^[1] The algorithm is implemented in all the major computer algebra systems.

Gosper-Petkovšek representation

Let ${\textstyle \mathbb {K} }$ be a field of characteristic zero. A nonzero sequence ${\textstyle y(n)}$ is called hypergeometric if the ratio of two consecutive terms is rational, i.e. ${\textstyle y(n+1)/y(n)\in \mathbb {K} (n)}$ . The Petkovšek algorithm uses as key concept that this rational function has a specific representation, namely the Gosper-Petkovšek normal form. Let ${\textstyle r(n)\in \mathbb {K} [n]}$ be a nonzero rational function. Then there exist monic polynomials ${\textstyle a,b,c\in \mathbb {K} [n]}$ and ${\textstyle 0\neq z\in \mathbb {K} }$ such that

$r(n)=z{\frac {a(n)}{b(n)}}{\frac {c(n+1)}{c(n)}}$

and

${\textstyle \gcd(a(n),b(n+k))=1}$ for every nonnegative integer ${\textstyle k\in \mathbb {N} }$ ,
${\textstyle \gcd(a(n),c(n))=1}$ and
${\textstyle \gcd(b(n),c(n+1))=1}$ .

This representation of ${\textstyle r(n)}$ is called Gosper-Petkovšek normal form. These polynomials can be computed explicitly. This construction of the representation is an essential part of Gosper's algorithm.^[2] Petkovšek added the conditions 2. and 3. of this representation which makes this normal form unique.^[1]

Algorithm

Using the Gosper-Petkovšek representation one can transform the original recurrence equation into a recurrence equation for a polynomial sequence ${\textstyle c(n)}$ . The other polynomials ${\textstyle a(n),b(n)}$ can be taken as the monic factors of the first coefficient polynomial ${\textstyle p_{0}(n)}$ resp. the last coefficient polynomial shifted ${\textstyle p_{r}(n-r+1)}$ . Then ${\textstyle z}$ has to fulfill a certain algebraic equation. Taking all the possible finitely many triples ${\textstyle (a(n),b(n),z)}$ and computing the corresponding polynomial solution of the transformed recurrence equation ${\textstyle c(n)}$ gives a hypergeometric solution if one exists.^[1]^[3]^[4]

In the following pseudocode the degree of a polynomial ${\textstyle p(n)\in \mathbb {K} [n]}$ is denoted by ${\textstyle \deg(p(n))}$ and the coefficient of ${\textstyle n^{d}}$ is denoted by ${\textstyle {\text{coeff}}(p(n),n^{d})}$ .

algorithm petkovsek is
    input: Linear recurrence equation  ${\textstyle \sum _{k=0}^{r}p_{k}(n)\,y(n+k)=0,p_{k}\in \mathbb {K} [n],p_{0},p_{r}\neq 0}$ .
    output: A hypergeometric solution  ${\textstyle y}$  if there are any hypergeometric solutions.

    for each monic divisor  ${\textstyle a(n)}$  of  ${\textstyle p_{0}(n)}$  do
        for each monic divisor  ${\textstyle b(n)}$  of  ${\textstyle p_{r}(n-r+1)}$  do
            for each  ${\textstyle k=0,\dots ,r}$  do
                 ${\textstyle {\tilde {p}}_{k}(n)=p_{k}(n)\prod _{j=0}^{k-1}a(n+j)\prod _{i=k}^{r-1}b(n+i)}$ 
         ${\textstyle d=\max _{k=0,\dots ,r}\deg({\tilde {p}}_{k}(n))}$ 
        for each root  ${\textstyle z}$  of  ${\textstyle \sum _{k=0}^{r}{\text{coeff}}({\tilde {p}}_{k}(n),n^{d})z^{k}}$  do
            Find non-zero polynomial solution  ${\textstyle c(n)}$  of  ${\textstyle \sum _{k=0}^{r}z^{k}\,{\tilde {p}}_{k}(n)\,c(n+k)=0}$ 
            if such a non-zero solution  ${\textstyle c(n)}$  exists then
                 ${\textstyle r(n)=z\,a(n)/b(n)\,c(n+1)/c(n)}$ 
                return a non-zero solution  ${\textstyle y(n)}$  of  ${\textstyle y(n+1)=r(n)\,y(n)}$

If one does not end if a solution is found it is possible to combine all hypergeometric solutions to get a general hypergeometric solution of the recurrence equation, i.e. a generating set for the kernel of the recurrence equation in the linear span of hypergeometric sequences.^[1]

Petkovšek also showed how the inhomogeneous problem can be solved. He considered the case where the right-hand side of the recurrence equation is a sum of hypergeometric sequences. After grouping together certain hypergeometric sequences of the right-hand side, for each of those groups a certain recurrence equation is solved for a rational solution. These rational solutions can be combined to get a particular solution of the inhomogeneous equation. Together with the general solution of the homogeneous problem this gives the general solution of the inhomogeneous problem.^[1]

Examples

Signed permutation matrices

The number of signed permutation matrices of size ${\textstyle n\times n}$ can be described by the sequence ${\textstyle y(n)}$ which is determined by the recurrence equation $4(n+1)^{2}\,y(n)+2\,y(n+1)+(-1)\,y(n+2)=0$ over ${\textstyle \mathbb {Q} }$ . Taking ${\textstyle a(n)=n+1,b(n)=1}$ as monic divisors of ${\textstyle p_{0}(n)=4(n+1)^{2},p_{2}(n)=-1}$ respectively, one gets ${\textstyle z=\pm 2}$ . For ${\textstyle z=2}$ the corresponding recurrence equation which is solved in Petkovšek's algorithm is $4(n+1)^{2}\,c(n)+4(n+1)\,c(n+1)-4(n+1)(n+2)\,c(n+2)=0.$ This recurrence equation has the polynomial solution ${\textstyle c(n)=c_{0}}$ for an arbitrary ${\textstyle c_{0}\in \mathbb {Q} }$ . Hence ${\textstyle r(n)=2(n+1)}$ and ${\textstyle y(n)=2^{n}\,n!}$ is a hypergeometric solution. In fact it is (up to a constant) the only hypergeometric solution and describes the number of signed permutation matrices.^[5]

Irrationality of $\zeta (3)$

Given the sum

a(n)=\sum _{k=0}^{n}{{\binom {n}{k}}^{2}{\binom {n+k}{k}}^{2}},

coming from Apéry's proof of the irrationality of $\zeta (3)$ , Zeilberger's algorithm computes the linear recurrence

(n+2)^{3}\,a(n+2)-(17n^{2}+51n+39)(2n+3)\,a(n+1)+(n+1)^{3}\,a(n)=0.

Given this recurrence, the algorithm does not return any hypergeometric solution, which proves that $a(n)$ does not simplify to a hypergeometric term.^[3]